Answer
Finding the integrating factor:
$\mu(t)=\exp(\int t/2\ dt)$
$\mu(t)=e^{t^2/4}$
Finding the y function:
$y(t)=1/\mu(t) [\int \mu(s)(1)ds+c]$
$y(t)=e^{-t^2/4}[\int e^{s^2/4}ds+c]$
$y(t)=e^{-t^2/4}\int e^{s^2/4}ds+ce^{-t^2/4}$
Applying the L'Hospital rule on the first term:
$\lim_{t\rightarrow +\infty} \dfrac{\int e^{s^2/4}ds}{e^{t^2/4}}=\lim_{t\rightarrow +\infty}\dfrac{e^{t^2/4}}{t/2\ e^{t^2/4}}=2/t\rightarrow0$
For the second term, the exponential goes to 1, so $lim_{t\rightarrow+\infty} y(t)=c$
Work Step by Step
Finding the integrating factor:
$\mu(t)=\exp(\int t/2\ dt)$
$\mu(t)=e^{t^2/4}$
Finding the y function:
$y(t)=1/\mu(t) [\int \mu(s)(1)ds+c]$
$y(t)=e^{-t^2/4}[\int e^{s^2/4}ds+c]$
$y(t)=e^{-t^2/4}\int e^{s^2/4}ds+ce^{-t^2/4}$
Applying the L'Hospital rule on the first term:
$\lim_{t\rightarrow +\infty} \dfrac{\int e^{s^2/4}ds}{e^{t^2/4}}=\lim_{t\rightarrow +\infty}\dfrac{e^{t^2/4}}{t/2\ e^{t^2/4}}=2/t\rightarrow0$
For the second term, the exponential goes to 1, so $lim_{t\rightarrow+\infty} y(t)=c$
